Problem: The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered. 

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x, y, r, where (x, y) are the coordinates of the stadium's center ( - 103 ≤ x, y ≤ 103), and r (1 ≤ r ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line. 

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

Examples

Input

0 0 10
60 0 10
30 30 10


Output

30.00000 0.00000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) : x(_x), y(_y) {}
  point operator+(const point &that) const {
    return point(x + that.x, y + that.y);
  }
  point operator-(const point &that) const {
    return point(x - that.x, y - that.y);
  }
  double operator*(const point &that) const { return x * that.y - that.x * y; }
  point operator*(const double &that) const {
    return point(x * that, y * that);
  }
  point normal() { return point(y, -x); }
  double length() { return hypot(x, y); }
  double polar() { return atan2(y, x); }
};
int sign(double x) {
  if (fabs(x) < eps) return 0;
  return x > 0 ? 1 : -1;
}
point intersection(const point &u1, const point &u2, const point &v1,
                   const point &v2) {
  return u1 + (u2 - u1) * (((u1 - v1) * (v1 - v2)) / ((u1 - u2) * (v1 - v2)));
}
struct data {
  char sta;
  point u, v, c;
  double r;
};
int totRet;
point c[3], tmpRet[2];
double r[3];
data tmp[2];
double disptop(const point &u, const point &v) {
  return hypot(u.x - v.x, u.y - v.y);
}
data get(int a, int b) {
  data ret;
  if (sign(r[a] - r[b]) == 0) {
    ret.sta = 'L';
    ret.u = (c[a] + c[b]) * 0.5;
    ret.v = ret.u + (c[a] - c[b]).normal();
  } else {
    ret.sta = 'C';
    ret.u = c[a] + (c[b] - c[a]) * (r[a] / (r[a] + r[b]));
    ret.v = c[b] + (c[a] - c[b]) * (r[b] / (r[b] - r[a]));
    ret.c = (ret.u + ret.v) * 0.5;
    ret.r = disptop(ret.c, ret.u);
  }
  return ret;
}
int intersectLinetoCir(point u, point v, point c, double r) {
  return sign(fabs((u - c) * (v - c) / disptop(u, v)) - r) <= 0;
}
void intersectionLinetoCir(point u, point v, point c, double r, point &ret1,
                           point &ret2) {
  point tmp = intersection(c, c + (u - v).normal(), u, v);
  double dis = disptop(tmp, c);
  dis = r * r - dis * dis;
  dis = dis < 0 ? 0 : sqrt(dis);
  dis /= disptop(u, v);
  ret1 = tmp + (u - v) * dis;
  ret2 = tmp - (u - v) * dis;
}
int intersectCirtoCir(point c1, double r1, point c2, double r2) {
  double tmp = disptop(c1, c2);
  return sign(tmp - (r1 + r2)) <= 0 && sign(tmp - fabs(r1 - r2)) >= 0;
}
void intersectionCirtoCir(point c1, double r1, point c2, double r2, point &ret1,
                          point &ret2) {
  double d = disptop(c1, c2);
  d = (1 + (r1 * r1 - r2 * r2) / d / d) * 0.5;
  point u = c1 + (c2 - c1) * d;
  point v = u + (c2 - c1).normal();
  intersectionLinetoCir(u, v, c1, r1, ret1, ret2);
}
void getIntersectionDatatoData(data a, data b) {
  point tmp1, tmp2;
  if (a.sta == 'L' && b.sta == 'L') {
    tmpRet[totRet++] = intersection(a.u, a.v, b.u, b.v);
  } else if (a.sta == 'C' && b.sta == 'C') {
    if (intersectCirtoCir(a.c, a.r, b.c, b.r)) {
      intersectionCirtoCir(a.c, a.r, b.c, b.r, tmp1, tmp2);
      tmpRet[totRet++] = tmp1;
      tmpRet[totRet++] = tmp2;
    }
  } else {
    if (a.sta == 'C') swap(a, b);
    if (intersectLinetoCir(a.u, a.v, b.c, b.r)) {
      intersectionLinetoCir(a.u, a.v, b.c, b.r, tmp1, tmp2);
      tmpRet[totRet++] = tmp1;
      tmpRet[totRet++] = tmp2;
    }
  }
}
int check(point u) {
  double R[3];
  for (int i = 0; i < 3; ++i) R[i] = disptop(u, c[i]);
  return !sign(r[0] * R[1] - r[1] * R[0]) && !sign(r[0] * R[2] - r[2] * R[0]);
}
void solve() {
  for (int i = 0; i < 3; ++i) {
    scanf("%lf%lf%lf", &c[i].x, &c[i].y, &r[i]);
  }
  tmp[0] = get(0, 1);
  tmp[1] = get(1, 2);
  totRet = 0;
  getIntersectionDatatoData(tmp[0], tmp[1]);
  int flag = false;
  point ans;
  double rate, rr;
  for (int i = 0; i < totRet; ++i)
    if (check(tmpRet[i])) {
      if (!flag) {
        ans = tmpRet[i];
        rate = disptop(ans, c[0]);
      } else {
        rr = disptop(tmpRet[i], c[0]);
        if (rr < rate) {
          rate = rr;
          ans = tmpRet[i];
        }
      }
      flag = true;
    }
  if (flag) printf("%.5f %.5f\n", ans.x, ans.y);
}
int main() {
  solve();
  return 0;
}